The purpose of this blog is to present the story behind Fermat's Last Theorem and Wiles' proof in a way accessible to the mathematical amateur.

Tuesday, May 17, 2005

Fermat's Last Theorem: n = 4

The easiest proof for Fermat's Last Theorem is the case n = 4. The proof for n=3 is a bit tricker. What is nice about this proof is that it arises quite naturally from the solution to Pythagorean Triples but it also proves Fermat's Theorem for all values n > 2 where n is even if we can prove Fermat's Last Theorem for all values n > 2 where n is odd. In fact, we can use the same time of reasoning to prove for all values n > 2 where n is not a prime if we can prove it true for all values n > 2 where n is prime.

Many textbooks say Fermat himself published this proof. This is not completely true. Fermat published a proof showing that a right triangle cannot have its area equal to a square. Fermat's proof does by implication show that there is no solution to n=4 but it is a bit more complicated than the proof that I am about to show.

Corollary for FLT n > 2: FLT is proven if FLT is proven for all cases where n is a prime.

The lesson learned from all this, is that we only need to proof that Fermat's Last Theorem is true for prime values of n.

If we prove that Fermat's Last Theorem is true for a given prime number, then it follows that it is true for any number which is divisible by that prime. For example, if we prove that there is no integer solution for x3 + y3 = z3, then we have likewise proven that there is no solution for (xi)3 + (yi)3 = (zi)3 = x3i + y3i = z3i.

This would also prove that x9 + y9 = z9 has no nontrivial integer solution.

i.e. You give as a corollary "Corollary for FLT n > 2 and even: There is no solution to:x^2n + y^2n; = z^2n; where xyz ≠ 0" .((why couldn't you have just said Corollary for FLT : There is no solution to:x^4n + y^4n = z^2n where xyz ≠ 0" as this is equal to (x^n)^4 + (y^n)^4 = (z^n)^2 ?))

It does not prove the case for n≡2mod4. (You say at the top of the page "it also proves Fermat's Theorem for all values n > 2 where n is even."

I reread your proof and it leaves out a number of steps (they may not be obvious to everyone). Shouldn't it complete a loop by showing the assumption x^4 + y^4 = z^2 leads to a smaller triple of the form a^4 + b^4 = c^2 not just a smaller triple of form a^2 + b^2 = c^2 ?

I have a number of proofs for various things to do with FLT and the Pythagorean triples, most of them are pretty short but rule out alot of possible solutions, if you want them posted let me know.

The proof is correct. If there are any general solutions, then there must be primitive solutions (see here).

In the proof, I show that there are no solutions for a primitive Pythagorean triple. Using the logic in the link above, it follows that if there is no primitive solution, then there cannot be a general solution.

Thanks for the proof! There are a few parts that took me a while to figure out though:

First, a and b must be relatively prime because p and q must be relatively prime, which is in turn because x^2, y^2, and z are relatively prime. (This may have been what Ralph was talking about: step 4 uses the parametrization for primitive Pythagorean triples without specifically noting that p, q, and y must be coprime.)

Then, after step 6, it should be noted that a and b must be squares, by the same theorem that step 6 uses, because a and b are relatively prime and ab is square.

Hi Larry,I'm happy to discover this blog.I hope you will help me to present another solution of Fermat last theorem, made in 1989. It is not a joke.The person who find it was not promoted, for unknown (by me) reasons.

it's about a proof witch was not officialy verified. The person who writed, published the resume of the proof in 1989 and died in 1990, in Romania.I believe he had not many instruments to communicate with those interested about his solution.The fact that Andrew Wiles has the only accepted solution prove only that it's notorious. He made a "show" around his work, in order to capt the attention. So, he had the neccessary support for the later acceptance.The other one liked another way to do that. Discreetly. Maybe too discreet, but he was afraid to talk to much, at those moment.Now I try to rebuild his solution, in his memory.It's not easy, because it's not my professional occupation right now. The author of this solution is Ion Melcu, and he published it in France, in french.

Good luck in putting together the solution. It is very sad when a person dies so early and it would be a great tribute if there was some mathematical insight that could be found in the work already done.

In fairness to Professor Wiles, his solution is brilliant. In my view, it is one of the mathematical highpoints of the last century.

I must have missed something, but I don't know what. Teh theorem states : If the area of a right-angled triangle were a square... . I have a right angled triangle with height 4 and base 2, it's area is 4*2/2 = 4 a square ! The same holds for 16*2/2 = 16 a square. What did I miss ?

I must have missed something, but I don't know what. Teh theorem states : If the area of a right-angled triangle were a square... . I have a right angled triangle with height 4 and base 2, it's area is 4*2/2 = 4 a square ! The same holds for 16*2/2 = 16 a square. What did I miss ?You are right; I wonder how Fermat missed that.

@Michael Ejercito : This can't be!!! There is an infinite set of examples. Therefore something must be wrong in the wording of this theorem. It's impossible that Fermat and the rest of the mathematical world overlooked it. Where is then the error in the proof of this theorem p^4 - q^4 = z^2 ?

@Larry,thanks but how is the phrasing in the beginning of your proof ? "In his life, Pierre de Fermat only left one proof in relation to number theory. He used his method of infinite descend to show that the area of a right triangle CANNOT BE SQUARE WITHIN THE DOMAIN OF WHOLE NUMBERS." There must be a problem in the wording of the theorem, I think.

Excuse a simple question, but with Fermat's theorem, if a solution were to be found, would the numbers xyz necessarily have to be the sides of a right triangle? Or, put another way, if you could prove the theorem is true for the values of the sides of all right triangles, would you have proved the entire theory?

I don't quite understand to what extent Fermat's theorem deals, if at all, with right triangles.

FLT is about the case where n >= 3. Right Triangles involve n=2. These are known as Pythagorean Triples and there are an infinite number of them. This solution was famously discussed by the Ancient Greek Diophantus.

Hi Larry,I have the following, most probably, philosophical, question: Since it is proven that the theorem of Fermat is true, it follows that any theorem that would claim the opposite is untrue.Since one of the characteristics of mathematics is that it consists of a number of statements which are 1. proven to be valid starting from a limited number of hypothesis (the so-called axiomas); 2. up to now not proven to be inconsistent with each other; is it right to state the following theorem :"The square root of 2 is a rational number is equivalent to the statement that the Theorem of Fermat is not true" ? You will immediately notice that I could have replaced the second part of my statement by any other theorem of which we know that it is true. In more general terms, is it true that, up to our knowledge in mathematics, each inconsistency in mathematics can be reduced to another known inconsistency ?Of course, an obvious answer would be: you are right, if you can prove it; but I feel that this answer is the same as the one Gödel gave about inconsistencies in mathematics. Just to have your feeling about the question. Thanks

Fermat's Last Theorem is true. This was proven by Andrew Wiles.to be more precise, he finished the proof by proving that semistable elliptical curves can not be modular. (It had already been shown that an integer solution to n>=3 would imply the existence of modular semistable elliptic curve.)

Hi Larry, Your correspondents want to discuss aspects of your proof of n = 4, but I would like to submit a proof for FLT which may have been Fermat's own version. After playing around with Diophantus' solution to Q8 Bk II around the early 1630's, Fermat knew that Pythagorean triples could be described algebraically in terms of a power 2 raised to a unit power by the equation (x^2)^1 + (y^2)^1 = (z^2)^1, when x,y,z are integers. From this he found the equation (x^n)^k + (y^n)^k = (z^n)^k described any power separated into two powers, when power n is raised to power k. Your first conjectural reconstruction of "The easiest proof for Fermat's Last Theorem is the case n = 4...." concerns Fermat's marginal note to Q29 Bk V of the Arithmeticorum annotated around the mid 1630's, saying that he had proved x^4 + y^4 = z^2 had no solutions. Because values of z^4 can be written as z^2, it followed that (x^4)^1 + (y^4)^1 = (z^4)^1 had no solutions. Fermat vigorously proved his theorem by applying the logic of mathematical proof introduced by the ancient Greeks: Let A=(x^4)^1+(y^4)^1=(z^4)^1Let B=(x^4)^k>2+(y^4)^k>2=(z^4)^k>2Let C=(x^k>2)^4+(y^k>2)^4=(z^k>2)^4Let D=(x^k>2)^1+(y^k>2)^1=(z^k>2)^1-As A had no solutions, and A implied B, B had no solutions. - As B equalled C, C had no solutions. - As D implied C, D had no solutions. Replace (k>2) with any integer >2 in equations B, C, and D, and FLT (D) had no solutions when n>2 and k=1. In the early 1650's, Fermat derived his second proof of n = 4 by proving Pythagorean triangles do not have square areas, as shown in your reconstruction. The lesson not learned by all later mathematicians from Euler to Wiles, and even now in 2010, is that Fermat proved his theorem true for all n>2 by extending the case for n=4, not by taking different approaches to odd prime values of n. Phil Cutmore

Regarding Phil's "proof" of FLT...I will recap Phil's argument here:Let A=(x^4)^1+(y^4)^1=(z^4)^1Let B=(x^4)^k>2+(y^4)^k>2=(z^4)^k>2Let C=(x^k>2)^4+(y^k>2)^4=(z^k>2)^4Let D=(x^k>2)^1+(y^k>2)^1=(z^k>2)^1As A had no solutions, and A implied B, B had no solutions. As B equalled C, C had no solutions. As D implied C, D had no solutions.

Rather than point out flaws, I will restate it with k>1 rather than k>2:

Let A=(x^4)^1+(y^4)^1=(z^4)^1Let B=(x^4)^k>1+(y^4)^k>1=(z^4)^k>1Let C=(x^k>1)^4+(y^k>1)^4=(z^k>1)^4Let D=(x^k>1)^1+(y^k>1)^1=(z^k>1)^1As A had no solutions, and A implied B, B had no solutions. As B equalled C, C had no solutions. As D implied C, D had no solutions.

This second version of the proof is equally valid, as I'm sure Phil will agree, since the implications are the same: A->B, B=C, and D->C. Yet, it proves something which is clearly false. Therefore, there must be a flaw in both versions of the proof.

With the help of "lebesgue", on the NRICH forum, I now understand that equation (2b) y^2=p^2-q^2 is enough to establish that q must be even. This is a consequence of y^2 being odd, and thus one more than a multiple of 4.

Larry, since your goal is for your proof to be accessible to amateur mathematicians, maybe a little comment just after step 2 is in order to the effect that p is odd and q is even as a consequence of the 2nd equation in step 2.

Also, you might point out at the beginning that x is chosen to be even, and y is chosen to be odd WLOG.

Also, if you were to number your equations, then some of the text can be made shorter and clearer.

An example of all of these changes can be found here: http://mathhelp.wikia.com/wiki/Proof_x%5E4_%2B_y%5E4_%3D_z%5E2_has_no_solutions

Hi Larry, in reply to Graeme, my/Fermat's version demonstrates why k > 2 has to be 3 minimum, which is proved by implication that D = (x^3)^1 + (y^3)^1 = (z^3)^1 has no solutions. To prove FLT all we do is let k be any integer > 2 (not > 1) in B and C, and by implication D will never have any solutions. I agree that no proof of FLT occurs when k > 1 is 2 minimum, because D reduces to Pythagoras theorem. It is explained below why this "flaw" is a paradox, which Fermat resolves when working through the details of his proof. My proof is part of a much larger conjectural reconstruction of FLT. To understand some of Fermat's possible thought processes when playing around with powers of numbers, including his "Eureka" moment "...which this marvellous proposition truly explains", the following should be added in my version after "...had no solutions". "Because values of z^4 can be written as z^2, when k = 1, (x^4)^1 + (y^4)^1 = (z^4)^1 has no solutions. It follows that the multiples of n = 4 also have no solutions, i.e. when n = 8, 12, 16, 20, ...etc. because each of these power can be written as a fourth power, i.e. 2^8 = 4^4, 3^12 = 27^4. Now, the rules of powers allow them to be manipulated in different ways. If integer a raised to power n is a^n, then a^n raised to power k is (a^n)^k. This can be expressed as (a^n.k)^1, or (a^k.n)^1, or (a^k)^n. This means that each power can be written as a product of two powers n.k. Therefore the n.k multiples of n = 4 are 4 = 4.1, 8 = 4.2, 12 = 4.3, 16 = 4.4, 20 = 4.5, ...etc. From here Fermat simply played around with a few of the lower powers and found no solutions were possible. He asked himself how and why Pythagorean triples existed, and found that PT rearranged to (z^2)^1 - (y^2)^1 – (x^2)^1 always gave a zero answer. When applied to higher powers he found triples that only came close to, but not zero, i.e. when n = 3, triples (7,6,5), (9,8,6), (12,10,9); when n = 4, triples (3,2,1) and (9,8,7); and when n = 5 and higher, triples (3,2,1). Surprisingly, this proof by 'scientific experiment' demonstrated that no power greater than squares could be separated into two powers. But this was not a 'vigorous' proof demanded by the mathematical community. With this in view, Fermat went on to list all the powers n and multiples k of n = 3, 5, 6,...etc. that had no solutions. He included the multiples of n = 2 because something did not seem right: n n.k multiples 4 4.1 4.2 4.3 4.4 4.5... 2 2.1? 2.2 2.3 2.4 2.5...3 3.1 3.2 3.3 3.4 3.5...5 5.1 5.2 5.3 5.4 5.5...6 6.1 6.2 6.3 6.4 6.5...etc He was right because there is a paradox. If all powers (n = 3,4,5,6,... etc., including n = 2) do not have solutions, then neither do their multiples (n.k). But the first multiple of n = 2, (2.1) has solutions (as proved by Pythagoras theorem) but the second, third etc multiples, 2.2, 2.3, 2.4,...etc. do not. He resolved this paradox by equating multiples 2.2 to 4.1, and 2.3 to 3.2... etc. all of which had no solutions. But it was something he had to be careful of later on. His "Eureka" moment came when he noticed that n.k: 12.1 is the lowest common multiple (LCM) of n.k: 4.3 and n.k: 3.4, This meant that as the first multiple of n = 12 and the third multiple of n = 4 had no solutions, then the fourth multiple of n = 3 also had no solutions. Then by implication the first multiple of n = 3 had no solutions. From the way powers can manipulated Fermat realized that as any k > 2 can be a multiple of n = 4, then it would imply that (x^k)^1 + (y^k)^1 = (z^k)^1 had no solutions, and did not need to calculate the LCM at all. Continue: "Fermat vigorously proved his theorem..."regards Phil

OK, Fermat proved that the equation raised to the fourth power could not generate whole solutions using his theory of limitless descent, but it was Andrew Wiles who proved the remainder of the theorem in 1986. His proof is final and superior to all others which attempt to prove the theorem. It shows that there are no valid integer solutions for x*y*z<> 0 for all equations in x^n+y^n=z^n for all values greater than 3 and truly verifies Fermat's theorem. If you wish to learn more, read his theorem yourselves.

Step ( 9 ) isn't clear for the infinite descent in step (8) as it is not explained how the smaller square - P^2, has the same properties.This is, I believe the missing bit :-In step (6) it states ab and a^2+b^2 are both squares. In the same way, since a and b are relatively prime, it follows that a and b are both squares. So a=Q^2 and b=R^2 for integers Q, R.Therefore Q^4 + R^4 = P^2This is what step (9) refers to. I hope this has helped clarify the solution.

I don't know what school of Mathematical logic you attended but your argument falls down on the last line, ie D implied C so D had no solutions is nonsense as the implication is the wrong way.C would have to imply D to prove that.